3.111 \(\int (c i+d i x) (A+B \log (e (\frac {a+b x}{c+d x})^n)) \, dx\)

Optimal. Leaf size=86 \[ \frac {i (c+d x)^2 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{2 d}-\frac {B i n (b c-a d)^2 \log (a+b x)}{2 b^2 d}-\frac {B i n x (b c-a d)}{2 b} \]

[Out]

-1/2*B*(-a*d+b*c)*i*n*x/b-1/2*B*(-a*d+b*c)^2*i*n*ln(b*x+a)/b^2/d+1/2*i*(d*x+c)^2*(A+B*ln(e*((b*x+a)/(d*x+c))^n
))/d

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Rubi [A]  time = 0.06, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {2525, 12, 43} \[ \frac {i (c+d x)^2 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{2 d}-\frac {B i n (b c-a d)^2 \log (a+b x)}{2 b^2 d}-\frac {B i n x (b c-a d)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[(c*i + d*i*x)*(A + B*Log[e*((a + b*x)/(c + d*x))^n]),x]

[Out]

-(B*(b*c - a*d)*i*n*x)/(2*b) - (B*(b*c - a*d)^2*i*n*Log[a + b*x])/(2*b^2*d) + (i*(c + d*x)^2*(A + B*Log[e*((a
+ b*x)/(c + d*x))^n]))/(2*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2525

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[((d + e*x)^(m
+ 1)*(a + b*Log[c*RFx^p])^n)/(e*(m + 1)), x] - Dist[(b*n*p)/(e*(m + 1)), Int[SimplifyIntegrand[((d + e*x)^(m +
1)*(a + b*Log[c*RFx^p])^(n - 1)*D[RFx, x])/RFx, x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && RationalFunc
tionQ[RFx, x] && IGtQ[n, 0] && (EqQ[n, 1] || IntegerQ[m]) && NeQ[m, -1]

Rubi steps

\begin {align*} \int (111 c+111 d x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx &=\frac {111 (c+d x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{2 d}-\frac {(B n) \int \frac {12321 (b c-a d) (c+d x)}{a+b x} \, dx}{222 d}\\ &=\frac {111 (c+d x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{2 d}-\frac {(111 B (b c-a d) n) \int \frac {c+d x}{a+b x} \, dx}{2 d}\\ &=\frac {111 (c+d x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{2 d}-\frac {(111 B (b c-a d) n) \int \left (\frac {d}{b}+\frac {b c-a d}{b (a+b x)}\right ) \, dx}{2 d}\\ &=-\frac {111 B (b c-a d) n x}{2 b}-\frac {111 B (b c-a d)^2 n \log (a+b x)}{2 b^2 d}+\frac {111 (c+d x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{2 d}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 74, normalized size = 0.86 \[ \frac {i \left ((c+d x)^2 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )-\frac {B n (b c-a d) ((b c-a d) \log (a+b x)+b d x)}{b^2}\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*i + d*i*x)*(A + B*Log[e*((a + b*x)/(c + d*x))^n]),x]

[Out]

(i*(-((B*(b*c - a*d)*n*(b*d*x + (b*c - a*d)*Log[a + b*x]))/b^2) + (c + d*x)^2*(A + B*Log[e*((a + b*x)/(c + d*x
))^n])))/(2*d)

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fricas [B]  time = 0.82, size = 162, normalized size = 1.88 \[ \frac {A b^{2} d^{2} i x^{2} - B b^{2} c^{2} i n \log \left (d x + c\right ) + {\left (2 \, B a b c d - B a^{2} d^{2}\right )} i n \log \left (b x + a\right ) + {\left (2 \, A b^{2} c d i - {\left (B b^{2} c d - B a b d^{2}\right )} i n\right )} x + {\left (B b^{2} d^{2} i x^{2} + 2 \, B b^{2} c d i x\right )} \log \relax (e) + {\left (B b^{2} d^{2} i n x^{2} + 2 \, B b^{2} c d i n x\right )} \log \left (\frac {b x + a}{d x + c}\right )}{2 \, b^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*i*x+c*i)*(A+B*log(e*((b*x+a)/(d*x+c))^n)),x, algorithm="fricas")

[Out]

1/2*(A*b^2*d^2*i*x^2 - B*b^2*c^2*i*n*log(d*x + c) + (2*B*a*b*c*d - B*a^2*d^2)*i*n*log(b*x + a) + (2*A*b^2*c*d*
i - (B*b^2*c*d - B*a*b*d^2)*i*n)*x + (B*b^2*d^2*i*x^2 + 2*B*b^2*c*d*i*x)*log(e) + (B*b^2*d^2*i*n*x^2 + 2*B*b^2
*c*d*i*n*x)*log((b*x + a)/(d*x + c)))/(b^2*d)

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giac [B]  time = 0.78, size = 572, normalized size = 6.65 \[ \frac {1}{2} \, {\left (\frac {{\left (B b^{3} c^{3} i n - 3 \, B a b^{2} c^{2} d i n + 3 \, B a^{2} b c d^{2} i n - B a^{3} d^{3} i n\right )} \log \left (\frac {b x + a}{d x + c}\right )}{b^{2} d - \frac {2 \, {\left (b x + a\right )} b d^{2}}{d x + c} + \frac {{\left (b x + a\right )}^{2} d^{3}}{{\left (d x + c\right )}^{2}}} - \frac {B b^{4} c^{3} i n - 3 \, B a b^{3} c^{2} d i n - \frac {{\left (b x + a\right )} B b^{3} c^{3} d i n}{d x + c} + 3 \, B a^{2} b^{2} c d^{2} i n + \frac {3 \, {\left (b x + a\right )} B a b^{2} c^{2} d^{2} i n}{d x + c} - B a^{3} b d^{3} i n - \frac {3 \, {\left (b x + a\right )} B a^{2} b c d^{3} i n}{d x + c} + \frac {{\left (b x + a\right )} B a^{3} d^{4} i n}{d x + c} - A b^{4} c^{3} i - B b^{4} c^{3} i + 3 \, A a b^{3} c^{2} d i + 3 \, B a b^{3} c^{2} d i - 3 \, A a^{2} b^{2} c d^{2} i - 3 \, B a^{2} b^{2} c d^{2} i + A a^{3} b d^{3} i + B a^{3} b d^{3} i}{b^{3} d - \frac {2 \, {\left (b x + a\right )} b^{2} d^{2}}{d x + c} + \frac {{\left (b x + a\right )}^{2} b d^{3}}{{\left (d x + c\right )}^{2}}} + \frac {{\left (B b^{3} c^{3} i n - 3 \, B a b^{2} c^{2} d i n + 3 \, B a^{2} b c d^{2} i n - B a^{3} d^{3} i n\right )} \log \left (-b + \frac {{\left (b x + a\right )} d}{d x + c}\right )}{b^{2} d} - \frac {{\left (B b^{3} c^{3} i n - 3 \, B a b^{2} c^{2} d i n + 3 \, B a^{2} b c d^{2} i n - B a^{3} d^{3} i n\right )} \log \left (\frac {b x + a}{d x + c}\right )}{b^{2} d}\right )} {\left (\frac {b c}{{\left (b c - a d\right )}^{2}} - \frac {a d}{{\left (b c - a d\right )}^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*i*x+c*i)*(A+B*log(e*((b*x+a)/(d*x+c))^n)),x, algorithm="giac")

[Out]

1/2*((B*b^3*c^3*i*n - 3*B*a*b^2*c^2*d*i*n + 3*B*a^2*b*c*d^2*i*n - B*a^3*d^3*i*n)*log((b*x + a)/(d*x + c))/(b^2
*d - 2*(b*x + a)*b*d^2/(d*x + c) + (b*x + a)^2*d^3/(d*x + c)^2) - (B*b^4*c^3*i*n - 3*B*a*b^3*c^2*d*i*n - (b*x
+ a)*B*b^3*c^3*d*i*n/(d*x + c) + 3*B*a^2*b^2*c*d^2*i*n + 3*(b*x + a)*B*a*b^2*c^2*d^2*i*n/(d*x + c) - B*a^3*b*d
^3*i*n - 3*(b*x + a)*B*a^2*b*c*d^3*i*n/(d*x + c) + (b*x + a)*B*a^3*d^4*i*n/(d*x + c) - A*b^4*c^3*i - B*b^4*c^3
*i + 3*A*a*b^3*c^2*d*i + 3*B*a*b^3*c^2*d*i - 3*A*a^2*b^2*c*d^2*i - 3*B*a^2*b^2*c*d^2*i + A*a^3*b*d^3*i + B*a^3
*b*d^3*i)/(b^3*d - 2*(b*x + a)*b^2*d^2/(d*x + c) + (b*x + a)^2*b*d^3/(d*x + c)^2) + (B*b^3*c^3*i*n - 3*B*a*b^2
*c^2*d*i*n + 3*B*a^2*b*c*d^2*i*n - B*a^3*d^3*i*n)*log(-b + (b*x + a)*d/(d*x + c))/(b^2*d) - (B*b^3*c^3*i*n - 3
*B*a*b^2*c^2*d*i*n + 3*B*a^2*b*c*d^2*i*n - B*a^3*d^3*i*n)*log((b*x + a)/(d*x + c))/(b^2*d))*(b*c/(b*c - a*d)^2
- a*d/(b*c - a*d)^2)

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maple [F]  time = 0.18, size = 0, normalized size = 0.00 \[ \int \left (d i x +c i \right ) \left (B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )+A \right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*i*x+c*i)*(B*ln(e*((b*x+a)/(d*x+c))^n)+A),x)

[Out]

int((d*i*x+c*i)*(B*ln(e*((b*x+a)/(d*x+c))^n)+A),x)

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maxima [A]  time = 1.20, size = 156, normalized size = 1.81 \[ \frac {1}{2} \, B d i x^{2} \log \left (e {\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n}\right ) + \frac {1}{2} \, A d i x^{2} - \frac {1}{2} \, B d i n {\left (\frac {a^{2} \log \left (b x + a\right )}{b^{2}} - \frac {c^{2} \log \left (d x + c\right )}{d^{2}} + \frac {{\left (b c - a d\right )} x}{b d}\right )} + B c i n {\left (\frac {a \log \left (b x + a\right )}{b} - \frac {c \log \left (d x + c\right )}{d}\right )} + B c i x \log \left (e {\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n}\right ) + A c i x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*i*x+c*i)*(A+B*log(e*((b*x+a)/(d*x+c))^n)),x, algorithm="maxima")

[Out]

1/2*B*d*i*x^2*log(e*(b*x/(d*x + c) + a/(d*x + c))^n) + 1/2*A*d*i*x^2 - 1/2*B*d*i*n*(a^2*log(b*x + a)/b^2 - c^2
*log(d*x + c)/d^2 + (b*c - a*d)*x/(b*d)) + B*c*i*n*(a*log(b*x + a)/b - c*log(d*x + c)/d) + B*c*i*x*log(e*(b*x/
(d*x + c) + a/(d*x + c))^n) + A*c*i*x

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mupad [B]  time = 4.34, size = 134, normalized size = 1.56 \[ x\,\left (\frac {i\,\left (2\,A\,a\,d+4\,A\,b\,c+B\,a\,d\,n-B\,b\,c\,n\right )}{2\,b}-\frac {A\,i\,\left (2\,a\,d+2\,b\,c\right )}{2\,b}\right )+\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )\,\left (\frac {B\,d\,i\,x^2}{2}+B\,c\,i\,x\right )-\frac {\ln \left (a+b\,x\right )\,\left (B\,a^2\,d\,i\,n-2\,B\,a\,b\,c\,i\,n\right )}{2\,b^2}+\frac {A\,d\,i\,x^2}{2}-\frac {B\,c^2\,i\,n\,\ln \left (c+d\,x\right )}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*i + d*i*x)*(A + B*log(e*((a + b*x)/(c + d*x))^n)),x)

[Out]

x*((i*(2*A*a*d + 4*A*b*c + B*a*d*n - B*b*c*n))/(2*b) - (A*i*(2*a*d + 2*b*c))/(2*b)) + log(e*((a + b*x)/(c + d*
x))^n)*((B*d*i*x^2)/2 + B*c*i*x) - (log(a + b*x)*(B*a^2*d*i*n - 2*B*a*b*c*i*n))/(2*b^2) + (A*d*i*x^2)/2 - (B*c
^2*i*n*log(c + d*x))/(2*d)

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sympy [A]  time = 39.91, size = 444, normalized size = 5.16 \[ \begin {cases} c i x \left (A + B \log {\left (e \left (\frac {a}{c}\right )^{n} \right )}\right ) & \text {for}\: b = 0 \wedge d = 0 \\A c i x + \frac {A d i x^{2}}{2} - \frac {B c^{2} i n \log {\left (c + d x \right )}}{2 d} + B c i n x \log {\relax (a )} - B c i n x \log {\left (c + d x \right )} + \frac {B c i n x}{2} + B c i x \log {\relax (e )} + \frac {B d i n x^{2} \log {\relax (a )}}{2} - \frac {B d i n x^{2} \log {\left (c + d x \right )}}{2} + \frac {B d i n x^{2}}{4} + \frac {B d i x^{2} \log {\relax (e )}}{2} & \text {for}\: b = 0 \\c i \left (A x + \frac {B a n \log {\left (\frac {a}{c} + \frac {b x}{c} \right )}}{b} + B n x \log {\left (\frac {a}{c} + \frac {b x}{c} \right )} - B n x + B x \log {\relax (e )}\right ) & \text {for}\: d = 0 \\A c i x + \frac {A d i x^{2}}{2} - \frac {B a^{2} d i n \log {\left (\frac {a}{c + d x} + \frac {b x}{c + d x} \right )}}{2 b^{2}} - \frac {B a^{2} d i n \log {\left (\frac {c}{d} + x \right )}}{2 b^{2}} + \frac {B a c i n \log {\left (\frac {a}{c + d x} + \frac {b x}{c + d x} \right )}}{b} + \frac {B a c i n \log {\left (\frac {c}{d} + x \right )}}{b} + \frac {B a d i n x}{2 b} - \frac {B c^{2} i n \log {\left (\frac {c}{d} + x \right )}}{2 d} + B c i n x \log {\left (\frac {a}{c + d x} + \frac {b x}{c + d x} \right )} - \frac {B c i n x}{2} + B c i x \log {\relax (e )} + \frac {B d i n x^{2} \log {\left (\frac {a}{c + d x} + \frac {b x}{c + d x} \right )}}{2} + \frac {B d i x^{2} \log {\relax (e )}}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*i*x+c*i)*(A+B*ln(e*((b*x+a)/(d*x+c))**n)),x)

[Out]

Piecewise((c*i*x*(A + B*log(e*(a/c)**n)), Eq(b, 0) & Eq(d, 0)), (A*c*i*x + A*d*i*x**2/2 - B*c**2*i*n*log(c + d
*x)/(2*d) + B*c*i*n*x*log(a) - B*c*i*n*x*log(c + d*x) + B*c*i*n*x/2 + B*c*i*x*log(e) + B*d*i*n*x**2*log(a)/2 -
B*d*i*n*x**2*log(c + d*x)/2 + B*d*i*n*x**2/4 + B*d*i*x**2*log(e)/2, Eq(b, 0)), (c*i*(A*x + B*a*n*log(a/c + b*
x/c)/b + B*n*x*log(a/c + b*x/c) - B*n*x + B*x*log(e)), Eq(d, 0)), (A*c*i*x + A*d*i*x**2/2 - B*a**2*d*i*n*log(a
/(c + d*x) + b*x/(c + d*x))/(2*b**2) - B*a**2*d*i*n*log(c/d + x)/(2*b**2) + B*a*c*i*n*log(a/(c + d*x) + b*x/(c
+ d*x))/b + B*a*c*i*n*log(c/d + x)/b + B*a*d*i*n*x/(2*b) - B*c**2*i*n*log(c/d + x)/(2*d) + B*c*i*n*x*log(a/(c
+ d*x) + b*x/(c + d*x)) - B*c*i*n*x/2 + B*c*i*x*log(e) + B*d*i*n*x**2*log(a/(c + d*x) + b*x/(c + d*x))/2 + B*
d*i*x**2*log(e)/2, True))

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